Monday, June 23, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 1 PHYS 1441 – Section 001 Lecture...
-
Upload
kody-siford -
Category
Documents
-
view
217 -
download
2
Transcript of Monday, June 23, 2014PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu 1 PHYS 1441 – Section 001 Lecture...
Monday, June 23, 2014 PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu
1
PHYS 1441 – Section 001Lecture #11
Monday, June 23, 2014Dr. Jaehoon Yu
• Newton’s Law of Universal Gravitation• Weightlessness• Work done by a constant force• Multiplication of Vectors• Work-Kinetic Energy Theorem
Announcements• Term exam #2
– In class this Wednesday, June 25– Non-comprehensive exam– Covers CH 4.7 to what we finish tomorrow, Tuesday, June 24– Bring your calculator but DO NOT input formula into it!
• Your phones or portable computers are NOT allowed as a replacement!– You can prepare a one 8.5x11.5 sheet (front and back) of
handwritten formulae and values of constants for the exam no solutions, derivations or definitions!• No additional formulae or values of constants will be provided!
Monday, June 23, 2014 PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu
2
Monday, June 23, 2014 PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu
Newton’s Law of Universal GravitationPeople have been very curious about the stars in the sky, making observations for a long~ time. The data people collected, however, have not been explained until Newton has discovered the law of gravitation.
Every object in the Universe attracts every other object with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
How would you write this law mathematically?
G is the universal gravitational constant, and its value is
This constant is not given by the theory but must be measured by experiments.
With G gF
Unit? 22 / kgmN
This form of forces is known as the inverse-square law, because the magnitude of the force is inversely proportional to the square of the distances between the objects.
1m 1 22
12
mm
r2m G
3
Monday, June 23, 2014 PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu
F
Ex. Gravitational AttractionWhat is the magnitude of the gravitational force that acts on each particle in the figure, assuming m1=12kg, m2=25kg, and r=1.2m?
11 2 22
12 kg 25 kg6.67 10 N m kg
1.2 m
81.4 10 N
1 22
mmGr
4
Monday, June 23, 2014 PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu
Why does the Moon orbit the Earth?
5
Monday, June 23, 2014 PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu
Gravitational Force and Weight
Since weight depends on the magnitude of gravitational acceleration, g, it varies depending on geographical location.
The attractive force exerted on an object by the Earth
Gravitational Force, Fg
Weight of an object with mass M is W
By measuring the forces one can determine masses. This is why you can measure mass using the spring scale.
Mg
What is the SI unit of weight? N
6
Monday, June 23, 2014 PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu
W
W
g
Gravitational Acceleration
2EM m
Gr
mg
2EM m
Gr
mg
2EMGr
What is the SI unit of g? m/s2
Gravitational acceleration at distance r from the center of the earth!
7
Monday, June 23, 2014 PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu
Gravitational force on the surface of the earth:
Magnitude of the gravitational acceleration on the surface of the Earth
g
29.80 m s
GF 2EM m
Gr
2E
E
M mGRmg
2E
E
MGR
8
Monday, June 23, 2014
Example for Universal GravitationUsing the fact that g=9.80m/s2 on the Earth’s surface, find the average density of the Earth.
g
Since the gravitational acceleration is
EM
Therefore the density of the Earth is
2E
E
R
MG
2111067.6
E
E
R
M
E
E
V
M
3
2
4E
E
R
GgR
EGR
g
4
3
33611
/1050.51037.61067.64
80.93mkg
Solving for ME
2ER g
G
gF 2E
E
R
mMG mg Solving for g
PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu
9
Monday, June 23, 2014 PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu
There is only one speed that a satellite can have if the satellite is to remain in an orbit with a fixed radius.
Satellite in Circular Orbits
cF
EGM
r
What acts as the centripetal force?The gravitational force of the earth pulling the satellite!
2EmM
Gr
2v
mr
v
2 EGMv
r
10
Monday, June 23, 2014 PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu
Determine the speed of the Hubble Space Telescope orbiting at a height of 598 km above the earth’s surface.
11 2 2 24
6 3
6.67 10 N m kg 5.98 10 kg
6.38 10 m 598 10 m
Ex. Orbital Speed of the Hubble Space Telescope
v
37.56 10 m s 16900mi h
11
Monday, June 23, 2014 PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu
2EGM rv
r T
3 22
E
rT
GM
Period of a Satellite in an Orbit
Speed of a satellite
Period of a satellite
22EGM r
r T
2 32 2
E
rT
GM
Square either
side and solve for T2
This is applicable to any satellite or even for planets and moons.
Kepler’s 3rd Law
12
Monday, June 23, 2014 PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu
Geo-synchronous Satellites
What period should these satellites have?
Satellite TV
Global Positioning System (GPS)
The same as the earth!! 24 hours
13
Monday, June 23, 2014 PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu
In each case, what is the weight recorded by the scale?
Ex. Apparent Weightlessness and Free Fall
0
14
0
Monday, June 23, 2014 PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu
At what speed must the surface of the space station move so that the astronaut experiences a push on his feet equal to his weight on earth? The radius is 1700 m.
Ex. Artificial Gravity
15
Monday, June 23, 2014 PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu
16
x
y
Work Done by a Constant ForceA meaningful work in physics is done only when the net forces exerted on an object changes the energy of the object.
M
Fθ Free Body
DiagramM
d
θ
Which force did the work?
W
Force
How much work did it do?
What does this mean? Physically meaningful work is done only by the component of the force along the movement of the object.
Unit? N m
Work is an energy transfer!!
cosFd θ (for Joule)J
Why? What kind? Scalar
Let’s think about the meaning of work!• A person is holding a grocery bag and
walking at a constant velocity.• Are his hands doing any work ON the bag?
– No– Why not?– Because the force hands exert on the bag, Fp,
is perpendicular to the displacement!!– This means that hands are not adding any
energy to the bag.• So what does this mean?
– In order for a force to perform any meaningful work, the energy of the object the force exerts on must change due to that force!!
• What happened to the person?– He spends his energy just to keep the bag up
but did not perform any work on the bag.Monday, June 23, 2014 PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu17
Monday, June 23, 2014 PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu
18
W cos 0
Work done by a constant force
s cos90
cos180
1
0
1
s
Monday, June 23, 2014 PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu
19
Scalar Product of Two Vectors • Product of magnitude of the two vectors and the cosine of the
angle between them
• Operation is commutative
• Operation follows the distribution law of multiplication
• How does scalar product look in terms of components?
x x y y z zA B i i A B j j A B k k
kkjjii
ikkjji• Scalar products of Unit Vectors 1 0
x xA B y yA B z zA B =0
cross terms
Monday, June 23, 2014 PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu
20
Example of Work by Scalar ProductA particle moving on the xy plane undergoes a displacement d=(2.0i+3.0j)m as a constant force F=(5.0i+2.0j) N acts on the particle.
a) Calculate the magnitude of the displacement and that of the force.
b) Calculate the work done by the force F.
W
jiji 0.20.50.30.2 )(166100.20.30.50.2 Jjjii
Y
X
d F
Can you do this using the magnitudes and the angle between d and F?
W
22yx dd m6.30.30.2 22
22yx FF N4.50.20.5 22
Monday, June 23, 2014 PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu
21
Ex. Pulling A Suitcase-on-WheelFind the work done by a 45.0N force in pulling the suitcase in the figure at an angle 50.0o for a distance s=75.0m.
Does work depend on mass of the object being worked on?
W
Yes
Why don’t I see the mass term in the work at all then?
It is reflected in the force. If an object has smaller mass, it would take less force to move it at the same acceleration than a heavier object. So it would take less work. Which makes perfect sense, doesn’t it?
Monday, June 23, 2014 PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu
22
Ex. 6.1 Work done on a crateA person pulls a 50kg crate 40m along a horizontal floor by a constant force Fp=100N, which acts at a 37o angle as shown in the figure. The floor is rough and exerts a friction force Ffr=50N. Determine (a) the work done by each force and (b) the net work done on the crate.
What are the forces exerting on the crate?
FG=-mg
So the net work on the crate
Work done on the crate by FG
Fp Ffr
Which force performs the work on the crate?
Fp Ffr
Work done on the crate by Fp:
Work done on the crate by Ffr:
This is the same as
FN=+mg
Work done on the crate by FN
Monday, June 23, 2014 PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu
23
cosF sθ
cos F sθ
W
Ex. Bench Pressing and The Concept of Negative Work
A weight lifter is bench-pressing a barbell whose weight is 710N a distance of 0.65m above his chest. Then he lowers it the same distance. The weight is raised and lowered at a constant velocity. Determine the work in the two cases.
What is the angle between the force and the displacement?
0 Fs
W 180 Fs
What does the negative work mean? The gravitational force does the work on the weight lifter!
Monday, June 23, 2014 PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu
24
The truck is accelerating at a rate of +1.50 m/s2. The mass of the crate is 120-kg and it does not slip. The magnitude of the displacement is 65 m. What is the total work done on the crate by all of the forces acting on it?
Ex. Accelerating a Crate
What are the forces acting in this motion?
Gravitational force on the crate, weight, W or Fg
Normal force force on the crate, FN
Static frictional force on the crate, fs
Monday, June 23, 2014 PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu
25
Ex. Continued…Lets figure out what the work done by each of the forces in this motion is.
Work done by the gravitational force on the crate, W or Fg
Work done by Normal force force on the crate, FN
Work done by the static frictional force on the crate, fs
W sf
W cos 90ogF s 0
W cos 90oNF s 0
ma 2120 kg 1.5m s 180N
sf s
Which force did the work? Static frictional force on the crate, fsHow? By holding on to the crate so that it moves with the truck!
Monday, June 23, 2014 PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu
26
Kinetic Energy and Work-Kinetic Energy Theorem• Some problems are hard to solve using Newton’s second law
– If forces exerting on an object during the motion are complicated– Relate the work done on the object by the net force to the change of the
speed of the object
MΣF
M
s
vi vf
Suppose net force ΣF was exerted on an object for displacement d to increase its speed from vi to vf.
The work on the object by the net force ΣF isW
2as as
W
Using the kinematic equation of motion
W
Work KE Kinetic Energy
Work Work done by the net force causes change in the object’s kinetic energy.
ma s 2 20
1
2 fm v v 2 20
1 1
2 2fmv mv
2 20
2fv v
2 21 1
2 2f imv mv f iKE KE KE
Work-Kinetic Energy Theorem
cos 0ma s ma s
21
2mv
2 20fv v
Monday, June 23, 2014 PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu
27
When a net external force by the jet engine does work on an object, the kinetic energy of the object changes according to
W
Work-Kinetic Energy Theorem
Monday, June 23, 2014 PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu
28
The mass of the space probe is 474-kg and its initial velocity is 275 m/s. If a 56.0-mN force acts on the probe parallel through a displacement of 2.42×109m, what is its final speed?
Ex. Deep Space 1
Solve for vf
Monday, June 23, 2014 PHYS 1441-001, Summer 2014 Dr. Jaehoon Yu
29
A satellite is moving about the earth in a circular orbit and an elliptical orbit. For these two orbits, determine whether the kinetic energy of the satellite changes during the motion.
Ex. Satellite Motion and Work By the Gravity
For a circular orbit
For an elliptical orbit
No change! Why not?
Gravitational force is the only external force but it is perpendicular to the displacement. So no work.
Changes! Why?Gravitational force is the only external force but its angle with respect to the displacement varies. So it performs work.